Tag: newton's laws of motion

Questions Related to newton's laws of motion

The backside of a truck is open and a box of 40kg is placed 5m away from the rear end.The coefficient of friction of the box with the surface of the truck is 0.15.The truck starts from rest with $2m/s^2$ acceleration.Calculate the distance covered by the truck when the box falls off

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. 20m

  2. 30m

  3. 40m

  4. 50m

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Correct Option: A

A jet water issues from a nozzel with a velocity of $20 m/s$ and it impinges normally on a flat plate moving away from it at $10 m/s$. If the cross-sectional area of the jet is $0.02 m^2$ and the density of water is taken as $1000 kg/m^3$, then the force developed on the plate will be

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. $ 10 N$

  2. $ 100 N$

  3. $1000 N$

  4. $2000 N$

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Correct Option: A
Explanation:

A jet water issues from a nozzel with a velocity$=20m/s$

It impinges normally on a flat plate moving away from it$=10m/s$
Cross-sectional area of the jet $=0.02m^2$
The density of water is taken$=1000kg/m^3$
The force developed on the plate will be
Hydrostatic force o the bottom surface of the tank will be
$F _b=\rho g(l\times b)\times h\F _b\rho g\times(1\times2)\times 2\ F _b=4\rho g(lbh)\rightarrow(1)$
Hydrostatic force on vertical surface will be
$F _v=\rho g(l\times h)\times\cfrac{h}{2}\F _v=\rho g(2\times 2)\cfrac{lh^2}{2}\F _v=2\rho g(lh^2)\rightarrow(2)$
Ratio will be
$\cfrac{F _b}{F _V}=\cfrac{4\rho g(lbh)}{2\rho g(lh^2)}\ \cfrac{F _b}{F _V}=1\F _b=1\times10\ \quad=10N$

A man drops an apple in the lift. He finds that the apple remains stationary and does not fall. The lift is:

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. Going down with constant speed

  2. Going up with constant speed

  3. Going down with constant acceleration

  4. Going up with constant acceleration

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Correct Option: C
Explanation:

As the apple is dropped, it is under free-fall meaning that the force of gravity is acting on it. With respect to the person inside the lift, the apple seems not to be falling Hence, the man and the lift must also be falling with the action of acceleration due to gravity i.e, a constant acceleration.

option - C is correct.

When a lift is going up with uniform acceleration, the apparent weight of a person is $W _{1}$
When the lift is stationary, his apparent weight is $W _{2}$
When the lift falls freely his apparent weight is $W _{3}$
When the lift is going down with uniform acceleration which is less than the acceleration due to gravity, his apparent weight is $W _{4}$
The increasing order of these four weights is

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. $W _{1},W _{3},W _{2},W _{4}$

  2. $W _{3},W _{4},W _{2},W _{1}$

  3. $W _{3},W _{2},W _{4},W _{1}$

  4. $W _{2},W _{3},W _{4},W _{1}$

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Correct Option: B
Explanation:

When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. $W _{1}>mg$
When lift is stationary, $W _{2}=mg$
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. $W _{3}=0$
When lift accelerates down, pseudo force acts upwards,  hence it decreases apparent weight $W _{4}<mg$, but also the acceleration is less than g, therefore $W _{4}=m(g-a)>0$

Hence, $W _{3}<W _{4}<W _{2}<W _{1}$

Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west and B moves from west to east. Which train exerts greater force on the track?

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. A

  2. B

  3. they will exert equal force

  4. The mass and the speed of each train must be known to reach a conclusion.

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Correct Option: A
Explanation:

Working in the frame of universe:

$mg-N=m{ \omega  }^{ 2 }R$
$N=mg-m{ \omega  }^{ 2 }R$
Earth moves from west to east.
Now, since the train B moves from west to east, its $\omega$ is greater than A. 
Hence, ${ N } _{ B }<{ N } _{ A }$

A lift of mass $1000 \ Kg$ which is moving with acceleration of $1 m/s^{-2}$ in upward direction, then the tension developed in string which is connected to lift is : -

physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

  1. $9,800 N$

  2. $10,800 N$

  3. $11,000 N$

  4. $10,000 N$

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Correct Option: B
Explanation:

The mass of the lift is $1000\;{\rm{kg}}$, the acceleration is $1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ and the acceleration due to gravity is $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

The tension developed in string is given as,

$T = mg + ma$

$ = 1000 \times 9.8 + 1000 \times 1$

$ = 10800\;{\rm{N}}$

Thus, the tension developed in the string is $10800\;{\rm{N}}$.